Random walk on infinite graph with zero probability of leaving subgraph












0












$begingroup$


Let $G = (V,E)$ be an graph which is locally finite (every vertex has only finitely many edges - but there may be infinitely many vertices), and connected. Let $X_n$ be a random walk on $G$ that follows the transition probabilities induced by the edge weights of $G$.



Is it possible to find a $G$ such that there exists a nonempty $U subseteq V$ such that if we let $W$ be the subgraph of $G$ induced by the vertices $V setminus U$, then $W$ is connected and we have that for any $w in W$, $lim_{n rightarrow infty}P_w(X_n in W | X_{n-1} in W) = 1$ ?



($P_w$ denotes that $X_0 = w$)



Edit: Made some mistakes earlier. Corrected to indicate that the walk starts in $W$ and that both $W$ and $G$ are connected.






probability graph-theory random-variables markov-chains random-walk







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you indicate by $P_u$ the stationary distribution? If not, that cannot hold for all $n$, right? (Consider $n^*$ to be the first time $X_n$ enters $W$. Then $X_{n^*+1}$ has by definition a non-zero chance to leave $W$, by "coming back whence it comes" -- or what did I miss?)
    $endgroup$
    – Clement C.
    Dec 1 '18 at 22:58












  • $begingroup$
    Sorry, I made a mistake $P_u$ is the probability with respect to starting the walk at $u in W$
    $endgroup$
    – jackson5
    Dec 1 '18 at 23:03










  • $begingroup$
    Then, you may require the graph to be connected, since otherwise the answer is trivially "yes" (take $U$ to be a non-empty connected component, and $W$ another.)
    $endgroup$
    – Clement C.
    Dec 1 '18 at 23:04










  • $begingroup$
    Yes, made both edits just now. Thanks so much!
    $endgroup$
    – jackson5
    Dec 1 '18 at 23:04










  • $begingroup$
    And if the graph is connected then the answer is trivially "no".
    $endgroup$
    – Did
    Dec 1 '18 at 23:05
















0












$begingroup$


Let $G = (V,E)$ be an graph which is locally finite (every vertex has only finitely many edges - but there may be infinitely many vertices), and connected. Let $X_n$ be a random walk on $G$ that follows the transition probabilities induced by the edge weights of $G$.



Is it possible to find a $G$ such that there exists a nonempty $U subseteq V$ such that if we let $W$ be the subgraph of $G$ induced by the vertices $V setminus U$, then $W$ is connected and we have that for any $w in W$, $lim_{n rightarrow infty}P_w(X_n in W | X_{n-1} in W) = 1$ ?



($P_w$ denotes that $X_0 = w$)



Edit: Made some mistakes earlier. Corrected to indicate that the walk starts in $W$ and that both $W$ and $G$ are connected.






probability graph-theory random-variables markov-chains random-walk







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you indicate by $P_u$ the stationary distribution? If not, that cannot hold for all $n$, right? (Consider $n^*$ to be the first time $X_n$ enters $W$. Then $X_{n^*+1}$ has by definition a non-zero chance to leave $W$, by "coming back whence it comes" -- or what did I miss?)
    $endgroup$
    – Clement C.
    Dec 1 '18 at 22:58












  • $begingroup$
    Sorry, I made a mistake $P_u$ is the probability with respect to starting the walk at $u in W$
    $endgroup$
    – jackson5
    Dec 1 '18 at 23:03










  • $begingroup$
    Then, you may require the graph to be connected, since otherwise the answer is trivially "yes" (take $U$ to be a non-empty connected component, and $W$ another.)
    $endgroup$
    – Clement C.
    Dec 1 '18 at 23:04










  • $begingroup$
    Yes, made both edits just now. Thanks so much!
    $endgroup$
    – jackson5
    Dec 1 '18 at 23:04










  • $begingroup$
    And if the graph is connected then the answer is trivially "no".
    $endgroup$
    – Did
    Dec 1 '18 at 23:05














0












0








0





$begingroup$


Let $G = (V,E)$ be an graph which is locally finite (every vertex has only finitely many edges - but there may be infinitely many vertices), and connected. Let $X_n$ be a random walk on $G$ that follows the transition probabilities induced by the edge weights of $G$.



Is it possible to find a $G$ such that there exists a nonempty $U subseteq V$ such that if we let $W$ be the subgraph of $G$ induced by the vertices $V setminus U$, then $W$ is connected and we have that for any $w in W$, $lim_{n rightarrow infty}P_w(X_n in W | X_{n-1} in W) = 1$ ?



($P_w$ denotes that $X_0 = w$)



Edit: Made some mistakes earlier. Corrected to indicate that the walk starts in $W$ and that both $W$ and $G$ are connected.






probability graph-theory random-variables markov-chains random-walk







share|cite|improve this question











$endgroup$




Let $G = (V,E)$ be an graph which is locally finite (every vertex has only finitely many edges - but there may be infinitely many vertices), and connected. Let $X_n$ be a random walk on $G$ that follows the transition probabilities induced by the edge weights of $G$.



Is it possible to find a $G$ such that there exists a nonempty $U subseteq V$ such that if we let $W$ be the subgraph of $G$ induced by the vertices $V setminus U$, then $W$ is connected and we have that for any $w in W$, $lim_{n rightarrow infty}P_w(X_n in W | X_{n-1} in W) = 1$ ?



($P_w$ denotes that $X_0 = w$)



Edit: Made some mistakes earlier. Corrected to indicate that the walk starts in $W$ and that both $W$ and $G$ are connected.






probability graph-theory random-variables markov-chains random-walk




probability graph-theory random-variables markov-chains random-walk






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 23:12







jackson5

















asked Dec 1 '18 at 22:49









jackson5jackson5

606512




606512












  • $begingroup$
    Do you indicate by $P_u$ the stationary distribution? If not, that cannot hold for all $n$, right? (Consider $n^*$ to be the first time $X_n$ enters $W$. Then $X_{n^*+1}$ has by definition a non-zero chance to leave $W$, by "coming back whence it comes" -- or what did I miss?)
    $endgroup$
    – Clement C.
    Dec 1 '18 at 22:58












  • $begingroup$
    Sorry, I made a mistake $P_u$ is the probability with respect to starting the walk at $u in W$
    $endgroup$
    – jackson5
    Dec 1 '18 at 23:03










  • $begingroup$
    Then, you may require the graph to be connected, since otherwise the answer is trivially "yes" (take $U$ to be a non-empty connected component, and $W$ another.)
    $endgroup$
    – Clement C.
    Dec 1 '18 at 23:04










  • $begingroup$
    Yes, made both edits just now. Thanks so much!
    $endgroup$
    – jackson5
    Dec 1 '18 at 23:04










  • $begingroup$
    And if the graph is connected then the answer is trivially "no".
    $endgroup$
    – Did
    Dec 1 '18 at 23:05


















  • $begingroup$
    Do you indicate by $P_u$ the stationary distribution? If not, that cannot hold for all $n$, right? (Consider $n^*$ to be the first time $X_n$ enters $W$. Then $X_{n^*+1}$ has by definition a non-zero chance to leave $W$, by "coming back whence it comes" -- or what did I miss?)
    $endgroup$
    – Clement C.
    Dec 1 '18 at 22:58












  • $begingroup$
    Sorry, I made a mistake $P_u$ is the probability with respect to starting the walk at $u in W$
    $endgroup$
    – jackson5
    Dec 1 '18 at 23:03










  • $begingroup$
    Then, you may require the graph to be connected, since otherwise the answer is trivially "yes" (take $U$ to be a non-empty connected component, and $W$ another.)
    $endgroup$
    – Clement C.
    Dec 1 '18 at 23:04










  • $begingroup$
    Yes, made both edits just now. Thanks so much!
    $endgroup$
    – jackson5
    Dec 1 '18 at 23:04










  • $begingroup$
    And if the graph is connected then the answer is trivially "no".
    $endgroup$
    – Did
    Dec 1 '18 at 23:05
















$begingroup$
Do you indicate by $P_u$ the stationary distribution? If not, that cannot hold for all $n$, right? (Consider $n^*$ to be the first time $X_n$ enters $W$. Then $X_{n^*+1}$ has by definition a non-zero chance to leave $W$, by "coming back whence it comes" -- or what did I miss?)
$endgroup$
– Clement C.
Dec 1 '18 at 22:58






$begingroup$
Do you indicate by $P_u$ the stationary distribution? If not, that cannot hold for all $n$, right? (Consider $n^*$ to be the first time $X_n$ enters $W$. Then $X_{n^*+1}$ has by definition a non-zero chance to leave $W$, by "coming back whence it comes" -- or what did I miss?)
$endgroup$
– Clement C.
Dec 1 '18 at 22:58














$begingroup$
Sorry, I made a mistake $P_u$ is the probability with respect to starting the walk at $u in W$
$endgroup$
– jackson5
Dec 1 '18 at 23:03




$begingroup$
Sorry, I made a mistake $P_u$ is the probability with respect to starting the walk at $u in W$
$endgroup$
– jackson5
Dec 1 '18 at 23:03












$begingroup$
Then, you may require the graph to be connected, since otherwise the answer is trivially "yes" (take $U$ to be a non-empty connected component, and $W$ another.)
$endgroup$
– Clement C.
Dec 1 '18 at 23:04




$begingroup$
Then, you may require the graph to be connected, since otherwise the answer is trivially "yes" (take $U$ to be a non-empty connected component, and $W$ another.)
$endgroup$
– Clement C.
Dec 1 '18 at 23:04












$begingroup$
Yes, made both edits just now. Thanks so much!
$endgroup$
– jackson5
Dec 1 '18 at 23:04




$begingroup$
Yes, made both edits just now. Thanks so much!
$endgroup$
– jackson5
Dec 1 '18 at 23:04












$begingroup$
And if the graph is connected then the answer is trivially "no".
$endgroup$
– Did
Dec 1 '18 at 23:05




$begingroup$
And if the graph is connected then the answer is trivially "no".
$endgroup$
– Did
Dec 1 '18 at 23:05










1 Answer
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$begingroup$

One way to get
$$
lim_{n rightarrow infty}P_w(X_n in W | X_{n-1} in W) = 1
$$
is to take a graph whose random walk Markov chain is transient.



For instance, let $G$ be an infinite binary tree, and let $W$ be one of the branches from the root. Then with probability $1$ the random walk only returns to the root a finite number of times, so in the limit as $n to infty$ the probability of the random walk leaving $W$ is $0$.






share|cite|improve this answer









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    $begingroup$

    One way to get
    $$
    lim_{n rightarrow infty}P_w(X_n in W | X_{n-1} in W) = 1
    $$
    is to take a graph whose random walk Markov chain is transient.



    For instance, let $G$ be an infinite binary tree, and let $W$ be one of the branches from the root. Then with probability $1$ the random walk only returns to the root a finite number of times, so in the limit as $n to infty$ the probability of the random walk leaving $W$ is $0$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      One way to get
      $$
      lim_{n rightarrow infty}P_w(X_n in W | X_{n-1} in W) = 1
      $$
      is to take a graph whose random walk Markov chain is transient.



      For instance, let $G$ be an infinite binary tree, and let $W$ be one of the branches from the root. Then with probability $1$ the random walk only returns to the root a finite number of times, so in the limit as $n to infty$ the probability of the random walk leaving $W$ is $0$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        One way to get
        $$
        lim_{n rightarrow infty}P_w(X_n in W | X_{n-1} in W) = 1
        $$
        is to take a graph whose random walk Markov chain is transient.



        For instance, let $G$ be an infinite binary tree, and let $W$ be one of the branches from the root. Then with probability $1$ the random walk only returns to the root a finite number of times, so in the limit as $n to infty$ the probability of the random walk leaving $W$ is $0$.






        share|cite|improve this answer









        $endgroup$



        One way to get
        $$
        lim_{n rightarrow infty}P_w(X_n in W | X_{n-1} in W) = 1
        $$
        is to take a graph whose random walk Markov chain is transient.



        For instance, let $G$ be an infinite binary tree, and let $W$ be one of the branches from the root. Then with probability $1$ the random walk only returns to the root a finite number of times, so in the limit as $n to infty$ the probability of the random walk leaving $W$ is $0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 '18 at 23:30









        Misha LavrovMisha Lavrov

        45.1k656107




        45.1k656107






























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